x3 + y3 = N
This is one of the classical number theory problems
related to elliptic curves.
We first point out that if N = p3 M then
clearly ( x / p , y / p ) is a solution of
x3 + y3 = M, so we assume
that N is cube-free.
Selmer conjectured that if N is a prime congruent to 4, 7,
8 modulo 9 then a solution exists. Noam Elkies proved the 4 and 7 cases.
One can prove by 3-descent that no solution exists if N is a prime congruent
to 2 or 5 mod 9. For N composite very little seems to have been proven.
We thus attack the problem in a computational way.
Let s = x + y, so
which, considered as a quadratic in x, has a rational solution
if the discriminant is a square, so
Multiply by 144N2/s4 and we get
if u = 12N / s = 12N / (x+y) and v = 12 N d / s2.
Now x3 + y3 = (x+y) (x2-xy+y2),
so x2 - x y + y2 = N/s.
From the quadratic,
x y = (s3-N) /
3s so
|
(x-y)2 = N / s -
(s3-N)
/ 3s = (4N-s3) / 3s = d2
/ 9s2 |
|
and thus v = 36 N (x-y) / (x+y).
These relations can be inverted to give
|
x = |
36N + v
6u |
y = |
36N - v
6u |
|
|
The curves EN have no torsion points for N > 2,
so any solution to the original problem must come from curves with rank
greater than 0. We use the Birch and Swinnerton-Dyer conjecture to predict
those curves with positive rank and also to predict the heights of those
curves with rank 1.
Quite often we make use of the 3-isogeny with the curve
where these curves often have points with heights one-third
of those on EN.
We recover (u,v) from the transformations
|
u = |
t3 + 64N2
t2 |
v = |
r (t3
- 128N2)
t3 |
|
|
We investigated the range 1 < = N < = 9999 and found
that the predicted heights of generators for EN or EN'
were within the range of our home-grown software.
The results computed are given in the following 5 files where
x = X/Z and y = Y/Z.
(a) 1
<= N <= 1999
(b) 2000
<= N <= 3999
(c) 4000
<= N <= 5999
(d) 6000
<= N <= 7999
(e) 8000
<= N <= 9999
We might extend this further but not at the moment.
File translated from TEX by TTH,
version 3.83.
Latest revision 10 Sep 2008, 09:21.